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Trigonometry, 2ADV T1 2020 HSC 22

The diagram shows a regular decagon (ten-sided shape with all sides equal and all interior angles equal). The decagon has centre `O`.
 

The perimeter of the shape is 80 cm.

By considering triangle `OAB`, calculate the area of the ten-sided shape. Give your answer in square centimetres correct to one decimal place.  (4 marks)

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`492.4\ text(cm²)`

Show Worked Solution

`angle AOB = 360/10 = 36^@`

`AB = 80/10 = 8\ text(cm)`

`DeltaAOB\ text(is made up of 2 identical right-angled triangles)`
 

`tan 18^@` `= 4/x`
`x` `= 4/(tan 18^@)`

 

`:.\ text(Area of decagon)` `= 20 xx 1/2 xx 4/(tan 18^@) xx 4`
  `= 492.429…`
  `= 492.4\ text(cm²  (to 1 d.p.))`

Trigonometry, 2ADV T1 2017 HSC 11e

In the diagram, `OAB` is a sector of the circle with centre `O` and radius 6 cm, where  `/_ AOB = 30^@`.


 

  1.  Find the exact value of the area of the triangle `OAB`.  (1 mark)

  2. Find the exact value of the area of the shaded segment.  (1 mark)

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  1.  `9\ text(cm²)`
  2. `3 pi – 9\ text(cm²)`
Show Worked Solution
i.   `text(Area)\ Delta OAB` `= 1/2 ab sin C`
    `= 1/2 xx 6^2 xx sin 30^@`
    `= 9\ text(cm²)`

 

ii.  `text(Area segment)` `= text(Area sector) – text(Area)\ Delta OAB`
    `= 30/360 xx pi xx 6^2 – 9`
    `= 3 pi – 9\ \ text(cm²)`

Trigonometry, 2ADV T1 2007 HSC 4c

 
An advertising logo is formed from two circles, which intersect as shown in the diagram.

The circles intersect at `A` and `B` and have centres at `O` and `C`.

The radius of the circle centred at `O` is 1 metre and the radius of the circle centred at `C` is `sqrt 3` metres. The length of `OC` is 2 metres.

  1. Use Pythagoras’ theorem to show that  `/_OAC = pi/2`.  (1 mark)

  2. Find  `/_ ACO`  and  `/_ AOC`.  (2 marks)

  3. Find the area of the quadrilateral  `AOBC`.  (1 mark)

  4. Find the area of the major sector  `ACB`.  (1 mark)

  5. Find the total area of the logo (the sum of all the shaded areas).  (2 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `/_ ACO = pi/6\ ,\ /_ AOC = pi/3`
  3. `sqrt 3\ \ text(m²)`
  4. `(5 pi)/2\ text(m²)`
  5. `((19 pi + 6 sqrt 3)/6)\ text(m²)`
Show Worked Solution

i.

`text(In)\ Delta AOC`

`AO^2 + AC^2` `= 1^2 + sqrt 3^2`
  `=1 + 3`
  `= 4`
  `= OC^2`

 
`:. Delta AOC\ \ text(is right-angled and)\ \ /_OAC = pi/2`

 

ii.  `sin\  /_ACO` `= 1/2`
`:. /_ACO` `= pi/6`
`sin\  /_AOC` `= sqrt 3/2`
`:. /_AOC` `= pi/3`

 

iii.  `text(Area)\ AOBC`

`= 2 xx text(Area)\ Delta AOC`

`= 2 xx 1/2 xx b xx h`

`= 2 xx 1/2 xx 1 xx sqrt 3`

`= sqrt 3\ \ text(m²)`

 

iv.  `/_ACB = pi/6 + pi/6 = pi/3`

`:. /_ACB\ text{(reflex)}` `= 2 pi – pi/3`
  `= (5 pi)/3`

 
`text(Area of major sector)\ ACB`

`= theta/(2 pi) xx pi r^2`

`= {(5 pi)/3}/(2 pi) xx pi(sqrt 3)^2`

`= (5 pi)/6 xx 3`

`= (5 pi)/2\ text(m²)`

 

v.  `/_AOB = pi/3 + pi/3 = (2 pi)/3`

`:. /_AOB\ text{(reflex)}` `= 2 pi – (2 pi)/3`
  `= (4 pi)/3`

`text(Area of major sector)\ AOB`

`= {(4 pi)/3}/(2 pi) xx pi xx 1^2`

`= (2 pi)/3\ text(m²)`

 

`:.\ text(Total area of the logo)`

`= (5 pi)/2 + (2 pi)/3 + text(Area)\ AOBC`

`= (15 pi + 4 pi)/6 + sqrt 3`

`= ((19 pi + 6 sqrt 3)/6)\ text(m²)`

Trigonometry, 2ADV T1 2006 HSC 4a

In the diagram, `ABCD` represents a garden. The sector  `BCD`  has centre `B` and  `/_DBC = (5 pi)/6`

The points `A, B` and `C` lie on a straight line and  `AB = AD = 3` metres.

Copy or trace the diagram into your writing booklet.

  1. Show that  `/_DAB = (2 pi)/3.`  (1 mark)

  2. Find the length of  `BD`.  (2 marks)

  3. Find the area of the garden  `ABCD`.  (2 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `3 sqrt 3\ \ text(m)`
  3. `(9 sqrt 3 + 45 pi) / 4\ \ text(m²)`
Show Worked Solution
i.   

`text(Show)\ /_DAB = (2 pi)/3`

`/_DBA` `= pi – (5 pi)/6\ \ \ text{(π radians in straight angle}\ ABC text{)}`
  `= pi/6\ text(radians)`

 
`:. /_BDA = pi/6\ text(radians)\ \ \ text{(base angles of isosceles}\ Delta ADB text{)}`

`:. /_DAB` `= pi – (pi/6 + pi/6)\ \ \ text{(angle sum of}\ Delta ADB text{)}`
  `= (2 pi)/3\  text(radians … as required)`

 

ii.  `text(Using the cosine rule:)`

`BD^2` `= AD^2 + AB^2 – 2 xx AD xx AB xx cos {:(2 pi)/3`
  `= 9 + 9 – (2 xx 3 xx 3 xx -0.5)`
  `= 27`
`:. BD` `= sqrt 27`
  `= 3 sqrt 3\ \ text(m)`

 

iii.  `text(Area of)\ Delta ADB` `= 1/2 ab sin C`
  `= 1/2 xx 3 xx 3 xx sin{:(2 pi)/3`
  `= 9/2 xx sqrt3/2`
  `= (9 sqrt 3)/4\ \ text(m²)`

 
`text(Area of sector)\ BCD`

`= {(5 pi)/6}/(2 pi) xx pi r^2`

`= (5 pi)/12 xx (3 sqrt 3)^2`

`= (45 pi)/4\ \ text(m²)`

 

`:.\ text(Area of garden)\ ABCD`

`= (9 sqrt 3)/4 + (45 pi)/4`

`= (9 sqrt 3 + 45 pi)/4\ \ text(m²)`

Trigonometry, 2ADV T1 2010 HSC 6b

The diagram shows a circle with centre `O` and radius 5 cm.

The length of the arc `PQ` is 9 cm. Lines drawn perpendicular to `OP` and `OQ` at `P` and `Q` respectively meet at  `T`.
 

  1. Find  `/_POQ`  in radians.   (1 mark)

  2. Prove that  `Delta OPT`  is congruent to  `Delta OQT`.   (2 marks)

  3. Find the area of the shaded region.   (2 marks)

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  1. `9/5\ text(radians)`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `6.3\ text(cm)\ \ \ text{(to 1 d.p.)}`
  4. `9.0\ text(cm²)\ \ \ text{(to 1 d.p.)}`
Show Worked Solution
i.    `text(Length of Arc)` `= r theta`
  `9` `= 5 xx /_POQ`
  `:.\ /_ POQ` `= 9/5\ text(radians)`

 

ii.

`text(Prove)\ Delta OPT ~= Delta OQT`

`OT\ text(is common)`

 MARKER’S COMMENT: Know the difference between the congruency proof of `RHS` and `SAS`. Incorrect identification will lose a mark.

`/_OPT = /_OQT = 90°\ \ \ text{(given)}`

`OP = OQ\ \ \ text{(radii)}`

`:.\ Delta OPT ~= Delta OQT\ \ \ text{(RHS)}`
 

iii.   
`/_POT ` `= 1/2 xx /_POQ\ \ \ text{(from part (ii))}`
  `=1/2 xx 9/5`
  `= 9/10\ text(radians)`
♦♦ Mean mark below 30%.
MARKER’S COMMENT: Many students struggled to work in radians. Make sure you understand this concept.
`tan /_ POT` `= (PT)/(OP)`
`tan (9/10)` `= (PT)/5`
`PT` `= 5 xx tan(9/10)`
  `=6.3007…`
  `=6.3\ text(cm)\ \ text{(to 1 d.p.)}`

 

iv.    `text(Shaded Area = Area)\ OQTP\ – text(Area Sector)\ OQP`
♦ Mean mark 35%.
`text(Area)\ OQTP` `= 2 xx text(Area)\ Delta OPT`
  `=2 xx 1/2 xx OP xx PT`
  `= 5 xx 6.3007`
  `~~ 31.503…`
  `~~31.5\ text(cm²)`

 

`text(Area Sector)\ OQP` ` = 1/2 r^2 theta`
  `= 1/2 xx 25 xx 9/5`
  `= 22.5\ text(cm²)`

 

`:.\ text(Shaded Area)` `= 31.503\ – 22.5`
  `=9.003…`
  `=9.0\ text(cm²)\ \ \ text{(to 1 d.p.)}`